Chapter 15

# Chapter 15 - Chapter 15 1 City has 10 macro-cells each cell...

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Unformatted text preview: Chapter 15 1. City has 10 macro-cells each cell has 100 users ∴ total number of users = 1000 Cells are of size 1 sqkm maximum distance traveled to traverse = √ 2km ∴ time = √ 2 30 = 169 . 7 s In the new setup number of cells = 10 5 microcells total number of users = 1000 × 100 2 users time = √ 2 × 10 30 × 10 3 =1.69s ∴ number of users increases by 10000 and handoff time reduces by 1 / 100 2. See Fig 1 (0,0) (0,1) v u (1,1) (1,0) 60 Figure 1: Problem 2 D 2 = ( j 20) 2 + ( i 20) 2- 2( j 20)( i 20)cos(2 π/ 3) ⇒ D = 2 a p i 2 + j 2 + ij = √ 3 R p i 2 + j 2 + ij 3. diamond shaped cells, R= 100m D min = 600 m D = 2 KR K = D 2 R = 600 2 × 100 = 3 N = K 2 = 9 (a) number of cells per cluster = N = 9 (b) number of channels per cell = total number/N = 450/9 = 50 4. (a) R=1km D=6km N = A cluster A cell = √ 3 D 2 / 2 3 √ 3 R 2 / 2 = 1 3 ( D/R ) 2 = 1 3 6 2 = 12 number of cells per cluster = N = 12 D SHADED CELLS USE SAME FREQUENCY Figure 2: Problem 3 (b) number of channels in each cell = 1200/12 = 100 (c) p i 2 + j 2 + ij = 2 √ 3 ⇒ i = 2 , j = 2 5. R=10m D=60m γ I = 2 γ = 4 M = 4 for diamond shaped cells SIR a = R- γ I MD- γ = R- 2 4 D- 4 = 32400 SIR b = R- 4 4 D- 4 = 324 SIR c = R- 2 4 D- 2 = 9 SIR a > SIR b > SIR c 6. γ = 2 BPSK P b = 10- 6 → P b = Q ( √ 2 γ b ) ⇒ γ b = SIR = 4 . 7534 B = 50 MHz each user100 KHz = B s SIR = 1 M ( D R ) γ M=6 for hexagonal cells a 1 = 0 . 167 a 2 = 3 N > 1 a 2 ‡ SIR a 1 · 2 /γ ⇒ N ≥ 9 . 4879 ∴ N = 10 C u = 50 7. G = 100 ξ = 1 λ = 1 . 5 With no sectorization SIR = 1 ξ 3 G ( N c- 1)(1 + λ ) = 4 . 7534 N c = b 26...
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Chapter 15 - Chapter 15 1 City has 10 macro-cells each cell...

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